Question

A body is thrown vertically upwards with a velocity of 28 m/s. Find the distance it will cover in 2 seconds

Answer 1

Given

initial velocity of the body = 28 m/s

total time = 2 seconds

total acceleration = -g = -10

To find

Distance covered in 2 seconds.

Formula

s = ut + 0.5 gt²

Calculations

s = $ut - \frac{1}{2} g {t}^{2} </p><p></p><p> = 28 \times 2 - \frac{1}{2} \times 10 \times 4 \\ \\ = 56 - 20 \\ \\ = 36 \: m$

Hence, it will cover a distance of 36 m.

Answer 2

Answer:

The distance it will cover in 2 seconds is  36m.

Explanation:

Using equation of motion for uniform acceleration,

$s= ut+\frac{1}{2} gt^{2}$

Considering g= -10$m/s^{2}$ (Negative because the body is thrown vertically upwards)

s= $28\times 2 +\frac{1}{2} \times 10\times 2^{2}$

=$56-20$

= 36 m

The correct answer is 36m.